Suppose that the probability of getting a job offer (from a specific company) is 0.16. Suppose that we observe empirically that given that an applicant for the job was in fact offered a job, the probability of an on campus interview was 0.93. Also, we observe empirically that given the group that were not offered a job the probability of an on campus interview was of 0.08. Given that a person got a campus interview, what is the probability that they get a job offer? (Answer to five decimal places).

0.68889

Step-by-step explanation:

This can be formulated as the following question:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

So

Given that a person got a campus interview, what is the probability that they get a job offer?

P(B) is the probability of getting a job offer. So

P(A/B) is the probability of getting a campus interview, given that they get a job offer. So P(A/B) = 0.93.

P(A) is the probability ofgetting a campus interview. Of those 16% who got a job, 93% had a campus interview. Of those 84% who did not get a job, 8% had a campus interview. So

Finally

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!